[wolffporter]

It could be deduced from dW=F.dS.

We know work equation: W=F.S.
If we differentiate this equation, we get dW=F.dS+S.dF, which is at odds with the correct expression dW=F.dS. This means the differential expression is more fundamental than the integral expression.

But why should dW be expressed as F.dS?

Is that to do with intensive property (F) which is non-integrable, and extensive property (S) which is integrable?

[juanrga]

It could be deduced from dW=F.dS.

We know work equation: W=F.S.
If we differentiate this equation, we get dW=F.dS+S.dF, which is at odds with the correct expression dW=F.dS. This means the differential expression is more fundamental than the integral expression.

But why should dW be expressed as F.dS?

Is that to do with intensive property (F) which is non-integrable, and extensive property (S) which is integrable?

The definition is



This definition follows because the term F dx appears when one studies changes in kinetic energy. Thus it is useful to give it a name.

(many physicists define work as -W, but that is only a convention).

The differential form is



The expression



is valid only when F is constant.

Since F = P A (A is area) and volume dV = AdX

$$ dW = - P dV /$$

[umangarora]

Work is defined as product of force and displacement caused by it. or W=F.S
dW=F.dS   P=F/A(Pressure is force per unit area) and Area X Height is Volume. A.S=V or AdS=dV
so dW=PAdV/A= PdV